Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct str {
  int v, nex;
} edge[2010000];
int fst[2010000], first[2010000], nex[2010000];
int lk[2010000], rk[2010000], size[2010000], dep[201000], fa[201000],
    st[101000][22];
int root[2010000];
int e, n, m, q, top, ans[210];
void make_edge(int a, int b) {
  edge[++e].nex = fst[a];
  fst[a] = e;
  edge[e].v = b;
  edge[++e].nex = fst[b];
  fst[b] = e;
  edge[e].v = a;
}
void build(int L, int R, int val, int pas, int u) {
  size[u] = size[pas] + 1;
  if (L == R) return;
  int mid = (L + R) >> 1;
  if (val <= mid) {
    lk[u] = ++top;
    rk[u] = rk[pas];
    build(L, mid, val, lk[pas], lk[u]);
  } else {
    rk[u] = ++top;
    lk[u] = lk[pas];
    build(mid + 1, R, val, rk[pas], rk[u]);
  }
}
void dfs(int u) {
  dep[u] = dep[fa[u]] + 1;
  int pas = root[fa[u]];
  for (int i = first[u]; i; i = nex[i]) {
    root[u] = ++top;
    build(1, m, i, pas, root[u]);
    pas = root[u];
  }
  root[u] = pas;
  for (int i = fst[u]; i; i = edge[i].nex) {
    int v = edge[i].v;
    if (fa[u] == v) continue;
    fa[v] = u;
    dfs(v);
  }
  return;
}
void prepare() {
  for (int i = 1; i <= n; i++) st[i][0] = fa[i];
  for (int j = 1; j <= 20; j++)
    for (int i = 1; i <= n; i++) st[i][j] = st[st[i][j - 1]][j - 1];
}
int lca(int a, int b) {
  if (dep[a] < dep[b]) swap(a, b);
  int i;
  for (i = 19; i >= 0 && !st[a][i]; i--)
    ;
  while (dep[a] != dep[b]) {
    for (; dep[st[a][i]] < dep[b] && i > 0; i--)
      ;
    a = st[a][i];
  }
  for (i = 19; i >= 0 && !st[a][i]; i--)
    ;
  while (a != b) {
    for (; i >= 0 && st[a][i] == st[b][i]; i--)
      ;
    if (i < 0)
      a = fa[a], b = fa[b];
    else
      a = st[a][i], b = st[b][i];
  }
  return a;
}
int ask(int L, int R, int a, int b, int c, int d, int k) {
  if (k > size[a] + size[b] - size[c] - size[d]) return 0;
  if (L == R) return L;
  int js = size[lk[a]] + size[lk[b]] - size[lk[c]] - size[lk[d]];
  int mid = (L + R) >> 1;
  if (js >= k)
    return ask(L, mid, lk[a], lk[b], lk[c], lk[d], k);
  else
    return ask(mid + 1, R, rk[a], rk[b], rk[c], rk[d], k - js);
}
int query(int a, int b, int k) {
  int c = lca(a, b), d = fa[c];
  return (ask(1, m, root[a], root[b], root[c], root[d], k));
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    make_edge(a, b);
  }
  for (int i = 1; i <= m; i++) {
    int a;
    scanf("%d", &a);
    nex[i] = first[a];
    first[a] = i;
  }
  dfs(1);
  prepare();
  for (; q; q--) {
    top = 0;
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    ans[0] = 0;
    int k;
    while (top < c && ((k = query(a, b, top + 1)) != 0)) ans[++top] = k;
    printf("%d ", top);
    for (int i = 1; i < top; i++) printf("%d ", ans[i]);
    if (top > 0)
      printf("%d\n", ans[top]);
    else
      putchar('\n');
  }
  return 0;
}